Problem: $\begin{cases}d(1)=13\\\\ d(n)=d(n-1)+17 \end{cases}$ Find the $4^{\text{th}}$ term in the sequence.
Answer: This is a recursive formula. It tells us that the first term is $13$ and that the common difference is $17$. $\begin{aligned} {d(1)}&=13 \\\\ {d(2)}&={d(1)}+17=30 \\\\ {d(3)}&={d(2)}+17=47 \\\\ {d(4)}&={d(3)}+17=64 \end{aligned}$ The $4^{\text{th}}$ term is $64$.